---
title: "LTI output of a WSS input is WSS"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:lti-wssp-output"
source: https://www.bctengineer.com/q/semester_6/communication_systems/lti-wssp-output
site: BCT Engineering
---

# LTI output of a WSS input is WSS

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2081 Baisakh (CS II), 6 marks
- Video: [DCLec09:Transmission of Random Process through an LTI System](https://www.youtube.com/watch?v=K2Eed98UhIo) — Zareena Nishanth
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/lti-wssp-output
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

Show that the output signal of LTI (linear time invariant) system is also a wide sense stationary process if the input to it is also a WSSP.

## Short answer

For a WSS input X(t) with mean m_X and autocorrelation R_X(τ) passing through a stable LTI system h(t), the output mean is constant m_X H(0). The output autocorrelation R_Y(τ) = h(τ)*h(-τ)*R_X(τ) depends only on τ, and the output PSD is S_Y(f) = |H(f)|² S_X(f). Since the mean is constant and autocorrelation depends only on τ, Y(t) is WSS.

## Answer

Let a WSS process $X(t)$ (mean $m_X$, autocorrelation $R_X(\tau)$, PSD $S_X(f)$) be applied to a stable LTI system with impulse response $h(t)$ and $H(f)$:

$$
Y(t)=\int_{-\infty}^{\infty}h(\alpha)X(t-\alpha)\,d\alpha
$$

```text
  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y
```

**1. Mean of output.** Expectation and integration can be interchanged:

$$
\begin{aligned}
E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\
&= m_X\,H(0) \quad(\text{a constant, independent of } t)
\end{aligned}
$$

**2. Autocorrelation of output.**

$$
\begin{aligned}
R_Y(t,t+\tau) &= E[Y(t)Y(t+\tau)] \\
&= E\left[\int h(\alpha)X(t-\alpha)d\alpha\int h(\beta)X(t+\tau-\beta)d\beta\right] \\
&= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\
&= \int\!\!\int h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta
\end{aligned}
$$

because $X$ is WSS, its autocorrelation depends only on the time difference $(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta$. The result depends only on $\tau$, not on $t$:

$$
R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)
$$

**3. Mean-square value** $E[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta$ is finite and constant.

Since the mean is constant and the autocorrelation depends only on $\tau$, **$Y(t)$ is also wide-sense stationary**.

**4. Output PSD.** Taking the Fourier transform of $R_Y(\tau)$ ($h(\tau)\to H(f)$, $h(-\tau)\to H^*(f)$):

$$
S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)
$$

and the output power is $P_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df$. The phase of $H(f)$ has no effect on the output PSD.

**Example.** White noise ($S_X=N_0/2$, WSS) through an RC low-pass filter, $\lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2}$:

$$
S_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2},\qquad R_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}
$$

$R_Y$ depends only on $\tau$, confirming that the output is WSS; its power $R_Y(0)=N_0/4RC$ is now finite.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A WSS process with mean m_X passes through an LTI system with transfer function H(f). What is E[Y(t)]?
   A. m_X h(0)
   B. m_X H(0), a constant
   C. m_X H(f), varying with frequency
   D. 0, because the filter removes the mean
   Answer: B. E[Y(t)] = m_X * integral of h(a) da = m_X H(0). It does not depend on t, so the first WSS condition holds. (Key point: Mean of the output)

2. Why does R_Y(t, t+tau) turn out to be independent of t?
   A. Because h(t) is causal
   B. Because X is WSS, so E[X(t-a)X(t+tau-b)] depends only on the time difference tau+a-b
   C. Because the mean of Y is zero
   D. Because the integral over a and b removes all t terms
   Answer: B. WSS input means R_X depends only on the time difference, which is tau+alpha-beta, so t drops out. (Key point: Why R_Y depends only on tau)

3. Which expression gives the output autocorrelation of the LTI system?
   A. R_Y(tau) = h(tau) * R_X(tau)
   B. R_Y(tau) = H(f) R_X(tau)
   C. R_Y(tau) = h(tau) * h(-tau) * R_X(tau)
   D. R_Y(tau) = h(tau) h(-tau) R_X(tau)
   Answer: C. The double integral of h(a)h(b)R_X(tau+a-b) is the convolution h(tau)*h(-tau)*R_X(tau). (Key point: Output autocorrelation)

4. What is the output PSD S_Y(f) and the role of the phase of H(f)?
   A. |H(f)|^2 S_X(f); phase has no effect
   B. H(f) S_X(f); phase shifts the PSD
   C. |H(f)| S_X(f); phase scales the power
   D. |H(f)|^2 S_X(f); phase changes the total power
   Answer: A. Fourier transforming R_Y gives H(f)H*(f)S_X(f) = |H|^2 S_X, so only the magnitude matters. (Key point: Output PSD)

## Related in Communication Systems

- [Random process; LTI output is WSS if input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arandom-process-lti-wssp) — Communication Systems, chapter 9
- [Stochastic process; WSS random signal through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astochastic-wss-through-lti) — Communication Systems, chapter 9, asked 2×
- [NEB; mean and AC of WSS signal through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aneb-wssp-lti-mean-ac) — Communication Systems, chapter 9
- [Ergodic process; WSS random signals through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aergodic-wss-through-lti) — Communication Systems, chapter 9, asked 2×
- [LTI system output in time domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Alti-output-time-domain) — Communication Systems, chapter 1
- [Parseval's theorem; output PSD of LTI system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aparseval-power-lti-psd) — Communication Systems, chapter 1

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